- #1
evinda
Gold Member
MHB
- 3,741
- 0
Hello! (Wave)For $A \in \mathbb{R}^{n \times n}$ with $A=A^{T}$ we set:
$$\lambda:= \max \{ \langle x, A x \rangle: ||x||_2=1\} \\ \mu:= \min \{ \langle x,A x \rangle: ||x||_2=1\}$$
Then for $x \in S_{||\cdot||_{2}}$ we have:
$$\langle x, Ax \rangle \leq |\langle x, A x \rangle| \leq ||x||_2 ||Ax||_2 \leq ||A|| ||x||_2^2=||A||$$
i.e. $\sup_{||x||_2=1} ||Ax||_2 \leq ||A||$
and since $\sup_{||x||_2=1} |\langle x, Ax \rangle|= \max \{ |\lambda|, |\mu|\}$ we have that $\max \{ |\lambda|, |\mu|\} \leq ||A||$.
Theorem
If $A \in \mathbb{R}^{n \times n}$ symmetric matrix then
$$||A||= \sup \{ ||Ax||_2: ||x||_2=1\}= \sup \{ |\langle x, Ax \rangle|: ||x||_2=1\}= \max \{ |\lambda|, |\mu|\}$$
Proof:
It remains to show that $||A|| \leq \sup \{ |\langle x, Ax \rangle|: ||x||_2=1\}:=a$
$$||Ax||_2^2=\frac{1}{4} (4 ||Ax||_2^2)=\frac{1}{4}(4 \langle Ax, Ax \rangle)=\frac{1}{4} ( 2 \langle Ax,Ax \rangle+2 \langle A^2 x, x \rangle) \overset{\text{ for an arbritary } \gamma>0}{=} \frac{1}{4}(\langle A(\gamma x+ \gamma^{-1} Ax), \gamma x+ \gamma^{-1} Ax \rangle- \langle A(\gamma x- \gamma^{-1}Ax), \gamma x-\gamma^{-1}Ax) \leq \frac{1}{4}(a ||\gamma x+ \gamma^{-1}Ax||^2+a ||\gamma x-\gamma^{-1}Ax||^2) \leq \frac{a}{2}(||x||_2^2 \gamma^2 + ||Ax||_2^2 \frac{1}{\gamma^2})$$
We set $f(\gamma)=||x||_2^2 \gamma^2 + ||Ax||_2^2 \frac{1}{\gamma^2})$ and check where $f$ achieves its minimum.
$$\frac{1}{4} ( 2 \langle Ax,Ax \rangle+2 \langle A^2 x, x \rangle) \overset{\text{ for an arbritary } \gamma>0}{=} \frac{1}{4}(\langle A(\gamma x+ \gamma^{-1} Ax), \gamma x+ \gamma^{-1} Ax \rangle- \langle A(\gamma x- \gamma^{-1}Ax), \gamma x-\gamma^{-1}Ax) $$
$$\lambda:= \max \{ \langle x, A x \rangle: ||x||_2=1\} \\ \mu:= \min \{ \langle x,A x \rangle: ||x||_2=1\}$$
Then for $x \in S_{||\cdot||_{2}}$ we have:
$$\langle x, Ax \rangle \leq |\langle x, A x \rangle| \leq ||x||_2 ||Ax||_2 \leq ||A|| ||x||_2^2=||A||$$
i.e. $\sup_{||x||_2=1} ||Ax||_2 \leq ||A||$
and since $\sup_{||x||_2=1} |\langle x, Ax \rangle|= \max \{ |\lambda|, |\mu|\}$ we have that $\max \{ |\lambda|, |\mu|\} \leq ||A||$.
- Why does it hold that $\sup_{||x||_2=1} |\langle x, Ax \rangle|= \max \{ |\lambda|, |\mu|\}$ ?
Theorem
If $A \in \mathbb{R}^{n \times n}$ symmetric matrix then
$$||A||= \sup \{ ||Ax||_2: ||x||_2=1\}= \sup \{ |\langle x, Ax \rangle|: ||x||_2=1\}= \max \{ |\lambda|, |\mu|\}$$
Proof:
It remains to show that $||A|| \leq \sup \{ |\langle x, Ax \rangle|: ||x||_2=1\}:=a$
$$||Ax||_2^2=\frac{1}{4} (4 ||Ax||_2^2)=\frac{1}{4}(4 \langle Ax, Ax \rangle)=\frac{1}{4} ( 2 \langle Ax,Ax \rangle+2 \langle A^2 x, x \rangle) \overset{\text{ for an arbritary } \gamma>0}{=} \frac{1}{4}(\langle A(\gamma x+ \gamma^{-1} Ax), \gamma x+ \gamma^{-1} Ax \rangle- \langle A(\gamma x- \gamma^{-1}Ax), \gamma x-\gamma^{-1}Ax) \leq \frac{1}{4}(a ||\gamma x+ \gamma^{-1}Ax||^2+a ||\gamma x-\gamma^{-1}Ax||^2) \leq \frac{a}{2}(||x||_2^2 \gamma^2 + ||Ax||_2^2 \frac{1}{\gamma^2})$$
We set $f(\gamma)=||x||_2^2 \gamma^2 + ||Ax||_2^2 \frac{1}{\gamma^2})$ and check where $f$ achieves its minimum.
- Could you explain me why the following equality holds?
$$\frac{1}{4} ( 2 \langle Ax,Ax \rangle+2 \langle A^2 x, x \rangle) \overset{\text{ for an arbritary } \gamma>0}{=} \frac{1}{4}(\langle A(\gamma x+ \gamma^{-1} Ax), \gamma x+ \gamma^{-1} Ax \rangle- \langle A(\gamma x- \gamma^{-1}Ax), \gamma x-\gamma^{-1}Ax) $$